Let all varieties be smooth, projective over an arbitrary algebraically closed field $k$.
Assume $f:S\to C$ is a surjective morphism from a rational surface $S$ to a curve $C$.
I've just read in Schütt & Shioda's Mordell-Weil Lattices, beginning of Section $7.1$, that in this case we have
"$C=\Bbb{P}^1$ and $K=k(\Bbb{P}^1)=k(t)$ is a rational function field. This follows, for instance, from Lüroth’s theorem stating that the only function field $k(C)$ contained in a purely transcendental extension of $k$ is a rational function field."
Remark: the authors assume $f:S\to C$ is a relatively minimal elliptic fibration, but I don't think this makes any difference for the argument above.
From what I could find about it, Lüroth's theorem for dimension 2 and arbitrary characteristic works only if $k(S)/k(C)$ is a separable extension (see here, for example).
What am I missing?
It suffices to prove the following:
Theorem. Let $k$ be an algebraically closed field and $C$ an irreducible curve over $k$. If there is a dominant morphism $f\colon U\to C$ from an open subset $U\subset \mathbb{A}^n$ of an affine space, then $C$ is rational.
Proof. Fix a closed point $x_0\in U$. Since $f$ is dominant, there exists a closed point $x_1\in U$ such that $f(x_0)\neq f(x_1)$. Consider the straight line $L\subset \mathbb{A}^n$ passing through $x_0$ and $x_1$. Then $f|_{L\cap U}\colon L\cap U\to C$ is not constant and hence dominant. Since $L\cap U$ is rational, Lüroth's theorem for dimension 1 implies that $C$ is rational.